I came upon the following question:

A solid expands on heating because:

a) the potential energy of interaction between atoms in the solid is asymmetric about the equilibrium positions of atoms; or

b) the frequency of vibrations of atoms increases.

My attempt:

I did not make a huge deal out of the question, and simply thought that it is due to the increase in frequency of vibrations, which now causes greater oscillations about the lattice points, which in turn increases the average volume occupied by atoms, leading to expansion. However the answer is a).

Can someone explain, possibly using some model (like the spring-particle model of the lattice), why this is so, and why b) is NOT a cause of expansion?

  • 2
    $\begingroup$ If the potential is symmetric, all those thermal oscillations will have zero effect on the average atomic position. $\endgroup$
    – Jon Custer
    Commented Sep 13, 2016 at 13:04
  • 1
    $\begingroup$ So, for expansion, we dont have to consider the increase in average volume occupied by atoms, but only the shift of their average positions? $\endgroup$
    – Lelouch
    Commented Sep 13, 2016 at 14:48
  • $\begingroup$ In a symmetric potential, why would the average position change? $\endgroup$
    – Jon Custer
    Commented Sep 13, 2016 at 15:35
  • 3
    $\begingroup$ Good question. The usual explanation in school physics is b) : When a material gets hot it expands - this is because the molecules in it are moving about more vigorously and so need more room. There is an explanation here including discussion of the failure of the spring model. $\endgroup$ Commented Sep 14, 2016 at 22:10

1 Answer 1


A simple and naïve model* for thermal expansion has two neighboring atoms represented by a pair of balls connected by a spring.

enter image description here

Image source: General Physics (calculus based) Class Notes, Dr. Rakesh Kapoor from University of Alabama at Birmingham

Giving the balls more energy (increasing the temperature of the atoms) will increase the amplitude of the spring's oscillations, which increases the amount of space the atoms will take up. The frequency of oscillations—how quickly the balls move back and forth—is irrelevant.

As for what "asymmetry of atom equilibrium positions" has to do with it, take a look at the graph below. The equilibrium distance for these atoms (spring length in the ball model) is marked by the dotted line at 74 pm. Clearly the green potential energy line is not symmetric across this line. From the green line shape it can be seen that atoms at equilibrium distance require less energy input to pull apart than to compress by the same amount, explaining why the addition of energy typically results in an increase in distance rather than a decrease.

Hydrogen Bonding Diagram

Image source: Chemistry 301, University of Texas (online resource)

*From Thermal Expansion of Solids by C. Y. Ho and R. E. Taylor:

A popular explanation for thermal expansion assumes central forces between pairs of interacting atoms. The asymmetry of the potential energy well causes the mean distance between the atoms to increase when they vibrate along the line joining them. In many texts this is the only model given for thermal expansion... although crude and in some ways misleading for solids (it does not explain negative expansion), it reveals the most generally important mechanism for thermal expansion, and correctly suggests that atomic vibrations give rise to thermal expansion because of anharmonicity.

  • $\begingroup$ This seems like a good answer, why no upvotes? $\endgroup$ Commented Sep 30, 2016 at 23:08
  • $\begingroup$ Very nice answer. How though we show that indeed the average position will be shifted? For example, in a harmonic oscillator the average position is $$ \frac{\int_{0}^{T} x(t) dt} {\int_{0}^{T} dt} = 0$$. How do we know that for asymmetric potentials the above integral does not vanish? $\endgroup$ Commented Sep 9, 2022 at 19:32
  • $\begingroup$ @AntoniosSarikas the harmonic oscillator average position is zero exactly because it's symmetric. If it's not symmetric then it won't be zero. $\endgroup$
    – pentane
    Commented Sep 10, 2022 at 23:02
  • $\begingroup$ Agree with Suzu Hirose. But pull apart for one pair will be compression for the other. How then the molecule decides where to go? $\endgroup$
    – KeSHAW
    Commented Oct 6, 2023 at 20:15
  • $\begingroup$ @KeSHAW each atom pushes on all of its neighbors $\endgroup$
    – pentane
    Commented Oct 28, 2023 at 13:27

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